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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#124946 | #5648. Crossing the Railways | Hongzy# | WA | 1ms | 3608kb | C++17 | 5.4kb | 2023-07-15 19:54:19 | 2023-07-15 19:54:20 |
Judging History
answer
#include <bits/stdc++.h>
#define LOG(FMT...) fprintf(stderr, FMT);
#define rep(i, j, k) for(int i = j; i <= k; ++ i)
#define per(i, j, k) for(int i = j; i >= k; -- i)
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
using db = double;
using ll = long long;
using uint = unsigned int;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
mt19937 mt(chrono::system_clock::now().time_since_epoch().count());
uniform_int_distribution<ll> ran(0, 1ll << 62);
void ucin() { ios::sync_with_stdio(0); cin.tie(0); }
// uniform_real_distribution<double> dbran;
template<class T> inline void chkmax(T &x, const T &y) { if(x < y) x = y; }
template<class T> inline void chkmin(T &x, const T &y) { if(x > y) x = y; }
inline ll sqr(ll x) { return x * x; }
inline ll cub(ll x) { return x * x * x; }
const int N = 510;
const int INF = 1e9;
const db eps = 1e-5;
#define lt(x, y) ((x) < (y) - eps)
#define leq(x, y) ((x) < (y) + eps)
struct seg {
int r, a, b;
bool operator < (const seg &rhs) const {
if(r != rhs.r) return r < rhs.r;
return a < rhs.a;
}
} a[N];
int n, m, s, v, L[12], R[12];
int dp[N*2], X[N*2], Y[N*2];
bool inc(db l, db x, db r) {
return lt(l, x) && lt(x, r);
}
bool incLL(ll l, ll x, ll r) {
return l < x && x < r;
}
int find(int y, ll up) { //last <= y
if(!L[y]) return -1;
int l = L[y], r = R[y], ans = -1;
while(l <= r) {
int mid = (l + r) >> 1;
if(X[mid] <= up) l = (ans = mid) + 1;
else r = mid - 1;
}
return ans;
}
bool inc(db x1, int y1, db x2, int y2) {
// assert(lt(x1, x2) && y1 < y2);
db k = (x2 - x1) / (y2 - y1);
rep(i, y1, y2) {
db x0 = k * (i - y1) + x1;
int z = find(i, ll(x0));
if(z == -1) continue;
z = (z + 1) / 2;
if(inc(a[z].a, x0, a[z].b))
return true;
}
return false;
}
int ans;
void upd(int id, db k, int cur) {
rep(i, Y[id]+1, m+1) {
db x0 = k * (i - Y[id]) + X[id];
if(i == m+1) {
if(leq(x0, s)) {
ans = min(ans, cur);
}
break;
}
int z = find(i, ll(x0));
if(z == -1) continue;
int w = z + 1;
z = (z + 1) / 2;
if(inc(a[z].a, x0, a[z].b))
break;
if(Y[w] == i)
dp[w] = min(dp[w], cur + 1);
}
}
void DP(int id, int y0, db k, int cur) {
per(i, y0, 0) {
db x0 = k * (i - Y[id]) + X[id];
if(!i) {
if(leq(x0, 0))
cur = min(cur, 0);
break;
}
int z = find(i, ll(x0));
if(z == -1) continue;
int w = dp[z];
z = (z + 1) / 2;
if(inc(a[z].a, x0, a[z].b))
break;
cur = min(cur, w + 2);
}
dp[id] = min(dp[id], cur);
upd(id, k, cur);
}
int main() {
scanf("%d%d%d%d", &n, &m, &s, &v);
rep(i, 1, n) {
int a, b, r;
scanf("%d%d%d", &a, &b, &r);
::a[i] = {r, a, b};
}
sort(a + 1, a + n + 1);
rep(i, 1, 2*n) {
int y = a[(i + 1) / 2].r;
int x = i & 1 ? a[(i + 1) / 2].a : a[(i + 1) / 2].b;
X[i] = x, Y[i] = y;
if(!L[y]) L[y] = i;
R[y] = i;
// printf("(%d, %d)\n", x, y);
}
{ //start
db x0 = 0, x2 = (m + 1.0) * v;
if(!inc(x0, 0, x2, m + 1) && leq(x2, s)) {
puts("0");
cerr << "**\n";
return 0;
}
}
rep(i, 1, 2*n) dp[i] = INF;
//pass single point
rep(i, 1, 2*n) {
db x = X[i];
int y = Y[i];
db x2 = x + (m + 1.0 - y) * v;
db x0 = x - (db)y * v;
if(leq(0, x0) && leq(x2, s) && !inc(x0, 0, x2, m + 1)) {
puts("0");
cerr << "()\n";
return 0;
}
}
//pass two points
rep(i, 1, 2*n)
rep(j, i+1, 2*n)
if(X[i] < X[j] && Y[i] < Y[j] && (Y[j] - Y[i]) * v <= (X[j] - X[i])) {
// printf("(%d, %d)!\n", i, j);
db k = (Y[j] - Y[i]) / db(X[j] - X[i]);
db x0 = X[i] - Y[i] / k, x2 = X[j] + (m + 1 - Y[j]) / k;
if(leq(0, x0) && leq(x2, s) && !inc(x0, 0, x2, m + 1)) {
puts("0");
cerr << "[]\n";
return 0;
}
}
ans = INF;
rep(i, 1, 2*n) {
int x = X[i], y = Y[i];
db x0 = x - (db)y * v;
if(leq(0, x0) && !inc(x0, 0, x, y)) {
// printf("x0 = %.1f\n", x0);
dp[i] = 0;
upd(i, v, 0);
} else {
rep(j, 1, i-1) {
if(x > X[j] && y > Y[j] && (y - Y[j]) * v <= (x - X[j])) {
db v0 = (y - Y[j]) / db(x - X[j]);
db x0 = x - (db)y / v0;
if(leq(0, x0) && !inc(x0, 0, x, y)) {
dp[i] = 0;
upd(i, 1 / v0, 0);
break;
}
}
}
}
DP(i, y, v, INF);
rep(j, 1, i-1) {
if(x > X[j] && y > Y[j] && (y - Y[j]) * v <= (x - X[j])) {
if(!inc(X[j], Y[j], x, y)) {
dp[i] = min(dp[i], dp[j] + 1);
DP(i, Y[j], (x - X[j]) / db(y - Y[j]), dp[j] + 1);
}
}
}
if(dp[i] < INF) {
db x2 = x + (m + 1.0 - y) * v;
if(leq(x2, s) && !inc(x, y, x2, m + 1)) {
ans = min(ans, dp[i] + 1);
} else {
rep(j, i+1, 2*n) {
if(x < X[j] && y < Y[j] && (Y[j] - y) * v <= (X[j] - x)) {
db v0 = (Y[j] - y) / db(X[j] - x);
db x2 = x + (db)(m + 1 - y) / v0;
if(leq(x2, s) && !inc(x, y, x2, m + 1)) {
ans = min(ans, dp[i] + 1);
break;
}
}
}
}
}
}
if(ans == INF) ans = -1;
printf("%d\n", ans);
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3432kb
input:
4 3 5 1 1 2 1 3 4 1 2 3 2 3 4 3
output:
0
result:
ok single line: '0'
Test #2:
score: -100
Wrong Answer
time: 1ms
memory: 3608kb
input:
3 3 12 2 2 10 1 1 6 2 8 12 3
output:
0
result:
wrong answer 1st lines differ - expected: '2', found: '0'